Differentiation - Connected Rates of Change

Differentiation - Connected Rates of Change

TGM Original Questions

Water is poured into a conical container (as shown in the diagram) of base radius \(15\) cm and the semi-vertical angle of the cone is \(\theta\), where \(\sin \theta = 0.6\). The depth of water is \(h\) cm. Apply the inflow-rate model in parts (iii) and (iv) after filling has begun, when the water depth is positive.

  1. Find the total height of the container.
  2. The volume of water is \(V\) \(\text{cm}^3\). Show that \(V = 3\pi \left[ 500 - \frac{(20 - h)^3}{16} \right].\)
  3. Given that the water is poured at a rate of \(kh^2\) \(\text{cm}^3 / \text{s}\), where \(k\) is a constant, find the rate of increase of \(h\) in terms of \(\pi\), \(h\) and \(k\).
  4. Given that the rate of increase of \(h\) is \(\frac{64}{9\pi}\) cm/s when \(h = 10\) cm, find the value of \(k\).

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs a course plan.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs a course plan.

Answer:(i) \(20\text{ cm}\) (ii) \(V=3\pi\left[500-\dfrac{(20-h)^3}{16}\right]\) (iii) \(\dfrac{\mathrm{d}h}{\mathrm{d}t}=\dfrac{16kh^2}{9\pi(20-h)^2}\) (iv) \(k=4\)

Need help? Join our A-Maths tuition classes.

Learn more